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Physics - Rotation Question with Solution | TestHub

PhysicsRotationCalculation of M.I.Medium2 minPYQ_2024

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The radius of gyration of a solid sphere of mass 5 kg about $X Y$ is 5 m as shown in figure. The radius of the sphere is $\frac{5 x}{7} m$ , then the value of $x$ is:
NEET 2024 (Re-NEET)

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$I_{X Y} = I_{CM} + M R^{2} = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2} = \frac{7}{5} \times 5 R^{2} = 7 R^{2}$ ...(1) $I_{X Y} = M K^{2} = 5 \times 5^{2}$ ...(ii) $∴ 5 \times 5^{2} = 7 \times R^{2} [$ From (1) and (2)] $\Rightarrow R = \frac{5}{7} \times 5 = \frac{5 x}{7}$ (Given) $∴ x = 5$

Stream:NEETSubject:PhysicsTopic:RotationSubtopic:Calculation of M.I.

⏱ 2mℹ️ Source: PYQ_2024

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